What is the pOH of<br> a<br> 2.6 x 10-6 M H+ solution?

Is the formation of acid rain an exothermic or endothermic reaction?

Monica [59]

It is an exothermic reaction

4 0

3 years ago

Molar Volume of a gas at STP=22.4 L Example 3: Determine the volume of Carbon dioxide created when 15.0 grams of NaHCO3 are deco

Naily [24]

Answer:

20009 is answer is right 1345678899444

Explanation:

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4 0

3 years ago

You weighed out 0.020 g of your crude aspirin product in order to determine the amount of salicylic acid impurity. Following the

Kitty [74]

Answer:

The correct answer is 7.8 percent.

Explanation:

As mentioned in the given question, the absorbance (A) of the sample solution is 1.07. To find the concentration of aspirin, Beer's law is used, that is, A = ebc

Here, e is the extinction coefficient, which is equal to 139.322 M^-1cm^-1 as per the standard value for salicylic acid, b is the pathlength, which is equivalent to 1 cm. Now putting the values we get,

A = ebc

c = A / (eb)

c = 1.07 / (139.322 × 1)

c = 0.00768 M

Now to determine the percent salicylic acid in the sample, there is a need to compare the value of concentration determined with the concentration of aspirin given initially.

0.02 grams is the initial concentration of aspirin mentioned in the question. The molar mass of aspirin is 240 g/mol.

Therefore, the moles of aspirin will be,

0.02 / 240 = 8.33 × 10^-5 moles

The final volume of the diluted solution given is 10 ml or 0.01 liters.

The molarity of aspirin in the diluted solution will be,

c1 = 8.33 × 10^-5 / 0.01 = 8.33 × 10^-3 M or 0.00833 M

Now, the percent of salicylic acid in the product will be,

c1 - c / c1 × 100

(0.00833 - 0.00768) / 0.00833 × 100 = 7.8 %

5 0

3 years ago

A calorimeter holds 50 g water at 22.0°C. A sample of hot iron is added to the water. The final temperature of the water and iro

makkiz [27]

The change in enthalpy associated with the change in the water’s temperature is 1254 J.

<h3>What is specific heat?</h3>

The amount of heat required to increase the temperature of one gram of a substance by one Celsius degree is known as specific heat.

Enthalpy change will be calculated as:

ΔH = -cmΔT, where

m = mass of water = 50g

c = specific heat of water = 4.18J/g°C

ΔT = change in temperature = 28 - 22 = 6 °C

On putting values in the above equation, we get

ΔH = -(4.18)(50)(6) = -1254 J

Hence change in enthalpy of the reaction is -1254 J.

To know more about enthalpy change, visit the below link:

brainly.com/question/11628413

#SPJ1

7 0

2 years ago

Given the partial equation:

Nikolay [14]

Answer : The balanced chemical equation in acidic medium will be,

$IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in acidic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the less number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydrogen ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$IO_3^-(aq)+Sn^{2+}(aq)\rightarrow I^-(aq)+Sn^{4+}(aq)$

The oxidation-reduction half reaction will be :

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-$

First balance the main element in the reaction.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-$

Now balance oxygen atom on both side.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-\rightarrow I^-+3H_2O$

Now balance hydrogen atom on both side.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}$

Reduction : $IO_3^-+6H^+\rightarrow I^-+3H_2O$

Now balance the charge.

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}+2e^-$

Reduction : $IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O$

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : $2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-$

Reduction : $IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O$

The balanced chemical equation in acidic medium will be,

$IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)$

5 0

3 years ago

What is the pOH of a 2.6 x 10-6 M H+ solution?